Single Number

Single Number

 

Given an array of integers, every element appears twice except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

import java.util.Hashtable;
public class Solution {
    public int singleNumber(int[] A) {
        Hashtable<Integer, Integer> hashtable = new Hashtable<Integer,Integer>();
        int result = 0;
        
        for(int i = 0; i < A.length;i++){
            Integer element = hashtable.get(A[i]);
            if(null == element){
                hashtable.put(A[i], 1);//第一次放1
            }
            else if(1 == element){
                hashtable.put(A[i], 2);//第二次放2
            }
        }//for
        for(int i = 0; i < A.length; i++){
            if(1 == hashtable.get(A[i])){
                result = A[i];
                break;
            }
        }//for
        return result;
    }
}

 ps

因为题目有要求，用O（1）空间，O（n）时间复杂度，用了hash表后，空间复杂度变为了O(n)

这里可以用异或来处理

异或具有交换性，a ^ b = b ^ a

0 ^ a = a

将所有的A[i]异或起来，A[0] ^ A[1]....这样根据交换性，最后只剩下单个的个数

只扫描了一遍，时间复杂度为O(N),只用了一个存储单元空间复杂度为O(1)

参考：http://www.cnblogs.com/changchengxiao/p/3413294.html

public class Solution {
    public int singleNumber(int[] A) {
        int result = 0;
        for(int i = 0; i < A.length; i++){
            result ^= A[i];
        }
        
        return result;
    }
}